Problem: If $x \veebar y = 3x-8y$ and $x \star y = xy+4x-y$, find $(4 \star -2) \veebar 0$.
Solution: First, find $4 \star -2$ $ 4 \star -2 = (4)(-2)+(4)(4)-(-2)$ $ \hphantom{4 \star -2} = 10$ Now, find $10 \veebar 0$ $ 10 \veebar 0 = (3)(10)-(8)(0)$ $ \hphantom{10 \veebar 0} = 30$.